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Given a complex rational function $f$ and $z\in \mathbb C$, let $O^+(z)=\{f^n(z):n\geq 1\}$. Let $$D(f)=\big\{z\in \mathbb C:\overline{O^+(z)}=J(f)\big\}.$$

So $D(f)$ is the set of points whose (forward) orbits are dense in the Julia set $J(f)$. It is known that $D(f)$ is a dense $G_\delta$-subset of $J(f)$ (in particular it is uncountable).

An arc is any subset of $\mathbb C$ that is homeomorphic to the interval $[0,1]$.

Question. For every rational map $f$, is it true that $D(f)$ contains no arc?

Actually I do not even know the answer for polynomials. I suspect that it is easier answer in that context. Also I am interested in the case when $J(f)$ is locally connected.

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Even for polynomials, the answer is No: See the question on Smooth Julia Sets for links to papers with more details.

Note that for rational functions, there is an additional case, as sometimes the Julia set is the whole Riemann sphere. One chapter in my Ph.D. thesis contains a short survey of the work up to the early 90s on these. These are truly weird, as one can build a finite metric on the Riemann sphere from some of the examples, which happens to be singular at a dense number of points on the sphere!

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  • $\begingroup$ I think your answer is for the whole Julia set, and not the smaller set of points with dense orbits. Unless I misunderstood. Can you give an explicit function for which $D(f)$ contains an arc? $\endgroup$ Commented May 24, 2023 at 17:41
  • $\begingroup$ Yes, that's right. I guess D(f) would have lots of holes for the usual suspects (i.e. $z^2$ and $z^2-2$). Sorry! $\endgroup$ Commented May 25, 2023 at 21:05

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